Introduction to really hard math equation:
In this article we are going to discuss about the really hard math equation. In math hard equation is easy to solve the math problem. Mathematics is the learning of quantity, arrangement, space, and change. Math seeks out patterns that originate the new conjecture, and ascertain truth by precise deduction from properly selected axioms and definitions. Math is used throughout the whole world that has fundamental tool in various fields that include natural science, engineering, medicine, and the social sciences.
Example Problems for Really Hard Math Equation:
Really hard math equation – Example: 1
Minimize `J(y)=\int 2\pi y \sqrt{1+y'^2} dx`
Solution:
Our functional does not depend explicitly on x, so we can use the first integral `F - y' F_{y'} = c_1 .`
`y \sqrt{1+y'^2} - \frac{y y'^2}{\sqrt{1+y'^2}} = c_1`
`\frac{y}{\sqrt{1+y'^2}} = c_1`
`y' = \frac{\sqrt{y^2-c_2^2}}{c_2^2}`
`dx = \int \frac{c_2 dy}{\sqrt{y^2-c_1^2}}`
Letting `y = c_1 \cosh t, dy = c_1 \sinh t dt ,` we get:
`dx = \int \frac{c_1^2 \sinh t dt}{c_1 \sinh t} = \int c_1 dt`
`x = c_1 t + c_2`
`y = c_1` ` {cosh} \frac{x-c_2}{c_1}`
This problem is equivalent to finding the equation of a hanging cable of uniform density with fixed endpoints.
Really hard math equation – Example: 2
Determine the volume of the space below the paraboloid `x^2+y^2+z-4=0` and above the square in the xy-plane with vertices at `(0,0),(0,1),(1,0),(1,1).`
Solution:
The top surface is `z=4-x^2-y^2=f(x,y).`
The projection R of this surface on the xy plane into the square is the square itself with the given vertices. Thus
`V=\int int_R f(x,y)dx dy=\int_0^1\int_0^1(4-x^2-y^2)dx dy`
`=\int_0^1[4y-x^2y-\frac{y^3}{3}]_0^1 dx`
`=\int_0^1 [\frac{11}{3}-x^2]dx=[\frac{11}{3}x-\frac{x^3}{3}]_0^1=\frac{10}{3}`
Really hard math equation – Example: 3
Find the volume of the region bounded by the paraboloids `z=x^2+y^2` and `z=\frac{6-\frac{x^2+y^2}{2}}.`
Solution:
The curve of the intersection of the two paraboloids is given by `x^2+y^2=6-\frac{x^2+y^2}{2}` or `x^2+y^2=4`
Hence,the volume V is given by `V=\int int_R \int_{x^2+y^2}^{6-\frac{x^2+y^2}{2}} dz dy dx` where R is the region in the (x,y) plane bounded by `x^2+y^2=4`
Therefore, `V=\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}[6-\frac{1}{2}(x^2+y^2)-(x^2+y^2)]dy dx`
`=6\int_0^2\int_{0}^{\sqrt{4-x^2}}[4-(x^2+y^2)]dy dx`
`=6\int_{0}^{\frac{\pi}{2}}\int_0^2(4-r^2)r dr d\theta`
`=12\pi `
Practice Problems for Really Hard Math Equation:
1. Find the volume of the region in space bounded by the surface `z=1-(x^2+y^2)` on the sides by the planes `x=0,y=0,x+y=1` and below by the plane z=0.
`Answer: frac{1}{3}`
2. Minimize `J(y)=\int_0^1(1+y''^2)dx, y(0)=0,y'(0)=1,y(1)=1,y'(1)=1`
`Answer: y = x`
In this article we are going to discuss about the really hard math equation. In math hard equation is easy to solve the math problem. Mathematics is the learning of quantity, arrangement, space, and change. Math seeks out patterns that originate the new conjecture, and ascertain truth by precise deduction from properly selected axioms and definitions. Math is used throughout the whole world that has fundamental tool in various fields that include natural science, engineering, medicine, and the social sciences.
Example Problems for Really Hard Math Equation:
Really hard math equation – Example: 1
Minimize `J(y)=\int 2\pi y \sqrt{1+y'^2} dx`
Solution:
Our functional does not depend explicitly on x, so we can use the first integral `F - y' F_{y'} = c_1 .`
`y \sqrt{1+y'^2} - \frac{y y'^2}{\sqrt{1+y'^2}} = c_1`
`\frac{y}{\sqrt{1+y'^2}} = c_1`
`y' = \frac{\sqrt{y^2-c_2^2}}{c_2^2}`
`dx = \int \frac{c_2 dy}{\sqrt{y^2-c_1^2}}`
Letting `y = c_1 \cosh t, dy = c_1 \sinh t dt ,` we get:
`dx = \int \frac{c_1^2 \sinh t dt}{c_1 \sinh t} = \int c_1 dt`
`x = c_1 t + c_2`
`y = c_1` ` {cosh} \frac{x-c_2}{c_1}`
This problem is equivalent to finding the equation of a hanging cable of uniform density with fixed endpoints.
Really hard math equation – Example: 2
Determine the volume of the space below the paraboloid `x^2+y^2+z-4=0` and above the square in the xy-plane with vertices at `(0,0),(0,1),(1,0),(1,1).`
Solution:
The top surface is `z=4-x^2-y^2=f(x,y).`
The projection R of this surface on the xy plane into the square is the square itself with the given vertices. Thus
`V=\int int_R f(x,y)dx dy=\int_0^1\int_0^1(4-x^2-y^2)dx dy`
`=\int_0^1[4y-x^2y-\frac{y^3}{3}]_0^1 dx`
`=\int_0^1 [\frac{11}{3}-x^2]dx=[\frac{11}{3}x-\frac{x^3}{3}]_0^1=\frac{10}{3}`
Really hard math equation – Example: 3
Find the volume of the region bounded by the paraboloids `z=x^2+y^2` and `z=\frac{6-\frac{x^2+y^2}{2}}.`
Solution:
The curve of the intersection of the two paraboloids is given by `x^2+y^2=6-\frac{x^2+y^2}{2}` or `x^2+y^2=4`
Hence,the volume V is given by `V=\int int_R \int_{x^2+y^2}^{6-\frac{x^2+y^2}{2}} dz dy dx` where R is the region in the (x,y) plane bounded by `x^2+y^2=4`
Therefore, `V=\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}[6-\frac{1}{2}(x^2+y^2)-(x^2+y^2)]dy dx`
`=6\int_0^2\int_{0}^{\sqrt{4-x^2}}[4-(x^2+y^2)]dy dx`
`=6\int_{0}^{\frac{\pi}{2}}\int_0^2(4-r^2)r dr d\theta`
`=12\pi `
Practice Problems for Really Hard Math Equation:
1. Find the volume of the region in space bounded by the surface `z=1-(x^2+y^2)` on the sides by the planes `x=0,y=0,x+y=1` and below by the plane z=0.
`Answer: frac{1}{3}`
2. Minimize `J(y)=\int_0^1(1+y''^2)dx, y(0)=0,y'(0)=1,y(1)=1,y'(1)=1`
`Answer: y = x`
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